Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <iostream>
using namespace std;

int N; long double p[10009], dp[3009][3009];

int main() {
	cin >> N;
	for (int i = 1; i <= N; i++) cin >> p[i];

	dp[0][0] = 1;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j <= N; j++) {
			dp[i + 1][j + 1] += dp[i][j] * p[i + 1];
			dp[i + 1][j] += dp[i][j] * (1.0L - p[i + 1]);
		}
	}
	long double E = 0;
	for (int i = (N / 2) + 1; i <= N; i++) E += dp[N][i];
	printf("%.12Lf\n", E);
	return 0;
}